## Differential Equation - help

Forum for the GRE subject test in mathematics.
boo78
Posts: 36
Joined: Tue Oct 15, 2013 3:53 pm

### Differential Equation - help

This is an example from the PR Math GRE Review: dy/dx = x(x-2)/e^y. We tried to do it in the class I teach and ran into a problem. PR solves it by sep of var (duh), giving the solution e^y = x^3/3 - x^2 + c. Fine, but this would mean the domain is restricted to values of x for which the cubic is positive. The slope field, however, doesn't indicate any problems in these regions and seems to suggest there should still be solutions there. PR apparently attempts to solve this issue by placing an absolute value around the cubic, giving the general solution of y = log|x^3/3 - x^2 + c|, but when you graph this function it doesn't agree with the slope field in the regions where the cubic is negative. My class and I are thinking 1) that this is yet another typo in PR, and 2) that there must be some correct way to represent the full family of solutions indicated by the slope field. Thoughts? It's been way too many years since I took DiffEq and I can't find my text book.

redcar777
Posts: 34
Joined: Sun Nov 04, 2012 11:59 am

### Re: Differential Equation - help

I think its sort of a trick specific to log. Note that for real x != 0 (including x<0), the derivative of log( |x| ) is 1/x

boo78
Posts: 36
Joined: Tue Oct 15, 2013 3:53 pm

### Re: Differential Equation - help

Yeah, that's why I kept thinking the solution had something to do with absolute values, but it was obvious PR didn't do it right. Since posting, I've explored the slope field and overlaying solutions in more detail, and I'm now convinced that the solution does NOT involve absolute values at all. If you vary the value of c enough, you can see solutions that cover the entire slope field. Some of the things that make it tricky are that the solutions have vertical asymptotes that vary in position depending on c, which isn't obvious from either the differential equation or initial examination of the slope field, but is obvious when you look at the cubic argument of the log. Also, the solutions about the negative y-axis only appear if you graph c-values between 0 and 4/3, when the relative maximum of the cubic is positive but the relative minimum is still negative. So the answer is that y = ln ((1/3)x^3-x^2+c) is the full family of solutions, no absolute values needed. YES!

davidthedavid
Posts: 3
Joined: Wed Jan 08, 2014 11:50 pm

### Re: Differential Equation - help

I don't understand your reasoning, boo78. I think the answer y = ln|x^3/3 - x^2 + c| is actually correct.

Taking into account that d( ln|x| )/dx = 1/x, take dy/dx for y:

dy/dx = [ d(x^3/3 - x^2) + c)/dx ] / (x^3/3 - x^2 + c) [chain rule]
= (x^2 - 2x) / e^( ln|x^3/3-x^2 + c| )
= x(x-2) / e^y

I don't think you are correct that there exist points on the slope field where this isn't true.

boo78
Posts: 36
Joined: Tue Oct 15, 2013 3:53 pm

### Re: Differential Equation - help

Your error is in your substitution of e^y for x^3/3 - x^2 + c. e^y > 0, so this substitution is only valid when x^3/3 - x^2 + c > 0. Therefore the differential equation only has solutions for values of x and c that satisfy this inequality. Using the absolute value allows for x^3/3 - x^2 + c to be negative, and thus is not a valid solution in those regions. If you graph the slope field and y = ln |x^3/3 - x^2 + c|, you will observe this.

davidthedavid
Posts: 3
Joined: Wed Jan 08, 2014 11:50 pm

### Re: Differential Equation - help

Your statement that "e^y > 0" is only true on the set of real numbers. On the complex plane there are plenty of numbers y such that e^y < 0.

Does the problem state explicitly whether they are working in real or complex numbers?

Edit: I just plugged it in at "http://www.wolframalpha.com/input/?i=dy ... 2%29%2Fe^y" , and I think I might be wrong. The answer might not have the "absolute value" in it for the solution on the complex plane. As far as the real solution goes, you either need the absolute value or to restrict y to be above zero.